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Given an array of integers nums, find the minimum number of operations required to make all elements divisible by 3. In one operation, you can increase or decrease an element of the array by 1.

Clarifying Questions

  1. What is the range of the values in nums?
    • The values can be any integer within typical constraints for an array element in competitive programming.
  2. Is there a constraint on the length of the array nums?
    • The length of the array can be anything reasonable for LeetCode problems, typically up to 10^5.
  3. What kind of operations can we perform?
    • You can increase or decrease any element by 1 in a single operation.
  4. Are there any edge cases we need to handle, like empty arrays or arrays with a single element?
    • While the problem typically assumes a non-empty array, an empty array would trivially require 0 operations.

Strategy

  1. Understanding the Remainders:
    • For an element x, if x % 3 == 0, it is already divisible by 3.
    • If x % 3 == 1, we need 2 steps to make x divisible by 3 (either -1 or +2).
    • If x % 3 == 2, we need 1 step to make x divisible by 3 (either +1 or -2).
  2. Counting Each Case:
    • We traverse the array and count how many elements fall into each of the above three categories.
  3. Sum Up the Operations:
    • Calculate the total minimum operations required by summing the operations needed for each remainder group.
    • Total operations = count of x % 3 == 1 * 2 + count of x % 3 == 2.

Code

def min_operations_to_divisible_by_3(nums):
    remainder_1_count = 0
    remainder_2_count = 0
    
    for num in nums:
        remainder = num % 3
        if remainder == 1:
            remainder_1_count += 1
        elif remainder == 2:
            remainder_2_count += 1
            
    operations = remainder_1_count * 2 + remainder_2_count
    
    return operations

# Example usage
nums = [1, 2, 3, 4]
print(min_operations_to_divisible_by_3(nums))  # Output should be 3

Time Complexity

The time complexity of this solution is (O(n)), where (n) is the length of the array. This is because we are making a single pass over the array to count the remainders, and the operations required for each element are performed in constant time.

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